Nope :( The sequence starts with 1 because there is just one triangle with an area of 2 that satisfies the conditions. I ended up with a sequence similar to your previous one when I forgot to take the absolute value of area -- x1(y2-y3) + x2(y3-y1) + x3*(y1-y2))/2 . That could be a possibility in case you were using the same formula. In your first result - 1, 3, 6, 14.... , first two are right, third over-counted and rest are undercounted.